A more direct way to see this is to skip the Nernst-Einstein detour and use Stokes' law for the ion's terminal drift velocity under an electric field directly.
Under a field $E$, an ion of charge $ze$ feels an electric force $zeE$. In steady drift this force is balanced by the viscous drag from Stokes' law, $6\pi\eta r v$. Setting them equal:
\[ zeE = 6\pi\eta r v \quad\Rightarrow\quad u=\frac{v}{E}=\frac{ze}{6\pi\eta r} \]The elementary charge is the Faraday constant divided by Avogadro's number, $e=F/N_A$. Since $R=N_Ak$, $N_A=R/k$, so $e=Fk/R$, which folds back into the same $u=zFk/(6\pi\eta rR)$ formula, just built up from the drift-velocity picture instead of the diffusion-coefficient picture.
Put in the numbers for $z=1$, $F=96500\ \mathrm{C\,mol^{-1}}$, $k=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$, $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$:
\[ e=\frac{Fk}{R}=\frac{(96500)(1.38\times10^{-23})}{8.314}=1.602\times10^{-19}\ \mathrm{C} \]which is (as a sanity check) the textbook value of the electron charge, confirming the constants are being combined correctly.
\[ u=\frac{(1)(1.602\times10^{-19})}{6\pi(8.94\times10^{-4})(1.5\times10^{-10})}=\frac{1.602\times10^{-19}}{2.5277\times10^{-12}}=6.34\times10^{-8}\ \mathrm{m^2\,V^{-1}\,s^{-1}} \]So $y\approx6.34$ to $6.35$, matching the answer from the diffusion-coefficient route, since both are just two ways of writing the same physics.